Question

The range of values of $a$ for which the point $(a,4)$ is out side the circles $x^2+y^2+10x=0$ and $x^2+y^2+12x+20=0$ is

• A. $(-\infty ,-8)\cup (-2,6)\cup (6,+\infty )$
• B. $(-8,-2)$
• C. $(-\infty ,-8)\cup (-2,\infty )$
• D. none of these

Answer 1

The correct option is C $(-\infty ,-8)\cup (-2,\infty )$

Given, $[x^2+y^2+10x{]}_{a,4}>0$

$a^2+16+10a>0$
$a^2+10a+16>0$
$(a+8)(a+2)>0$
$a<-8$ and $x>-2$.

Therefore

$aϵ(-\infty ,8)\cup (-2,\infty )$

Also

$[x^2+y^2+12x+20{]}_{a,4}>0$

$a^2+16+12a+20>0$
$a^2+12a+36>0$
$(a+6{)}^2>0$

This is true for all $aϵR$.

Hence the solution set for 'a' is

$aϵ(-\infty ,-8)\cup (-2,\infty )$.